Why Symmetric-Strike Calls and Puts Can Have Different Prices
Summary
The note addresses whether Black–Scholes prices a call and a put equally when their strikes are equally far from the current underlying price in opposite directions. It distinguishes terminal payoffs from option values before expiry. At expiry, if the underlying finishes exactly midway between the two strikes, the two payoffs match; that equality does not imply equal prices earlier in the option’s life.
The explanation points to the underlying’s nonnegative price in the model: a put’s payoff is capped because the underlying cannot fall below zero, while a call’s payoff has no comparable upper bound. The possibility of large upward moves adds value to the call even if those states are unlikely. The discussion is qualitative and does not derive Black–Scholes prices or account for dividends, volatility skew, or other market features; its comparison concerns model prices under the stated setup.
Key ideas
- Equal payoffs at a particular terminal price do not establish equal option values before expiry.
- A put’s payoff is bounded by the underlying’s inability to fall below zero in the model.
- A call has unbounded upside, so possible large upward moves contribute value.
- The explanation is qualitative and addresses model pricing rather than observed market prices.
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Full text
# Equal prices for call and put options with symmetric strikes around contemporaneous price?
# Equal prices for call and put options with symmetric strikes around contemporaneous price?
Shouldn't (according to the Black-Scholes model) the price of a call option with a strike of an arbitrary amount away from the current asset's price, be equal to the price of a put option with the same "distance to exercise", just in the opposite direction? In other words: Shouldn't a call and put with symmetric strikes around the asset's value be priced equally? According to my calculation, they don't equate for neither ITM, nor OTM options. What's the reason for this inequality of option prices?
I'm aware that this does not apply to observed market prices. But shouldn't this hold for option prices computed with the traditional Black-Scholes OP model, as this formula assumes normally distributed returns?
## Answer by Kermittfrog (score 1, accepted)
https://quant.stackexchange.com/a/53678
Without delving into the mathematics of the problem, I think we can answer your question as follows.
Please note that subsequent thoughts hold for both, zero and positive interest rates.
- At expiry, your statement is correct: The payoff of a put option will be $(X_P-S)^+$ and the payoff of a call is $(S-X_C)^+$. If $X_C<X_P$, and if the underlying just happens to be equal to $\frac{X_P+X_S}{2}$, i.e. exactly between the two, both have the exact same value.
- At any time before expiry, the put option's payoff can be expected to be at most the difference between $X_P$ and $0$, i.e. the put option has a limited upside as the underlying process cannot go below zero (at least in the model). A call option, on the other hand has unlimited upside potential, i.e. the call payoff at expiry can be more than just $X_P-0$, e.g. if $S_T=X_C+kX_P$, $k>0$, then of course the call payoff will be $(kX_P)$ which may be larger than $X_P$. Although these states have a small probability, they nevertheless add to the present value of the call option. Hence, the call should always be 'a bit' more expensive for a given distance-to-strike, compared to the put option.
HTHShown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.